I'd like some help reinterpreting the asymptotic result on a pmf. The asymptotic result is a continuous function, whereas the mass function is discrete.
Say $P(Y(t)=k)=f(k,t)$. It is the case that $\lim_{t\rightarrow\infty}f(k,t)=0$. I want to know how the tail of the distribution decays for long times, so I let $k=e^{b t}$, with $0<b<1$. This gives \begin{equation} P(Y(t)=e^{b t})=g(t)+O(e^{-b t}) \end{equation} where $g(t)$ is the leading order term. For example take \begin{equation} P(Y(t)=k)=\frac{(1-e^{-t})^{k}}{k t} \end{equation} then making the substitution we have \begin{gather} P(Y(t)=e^{b t})=e^{-bt}t^{-1}+O(e^{-t}t^{-1}). \end{gather} What can we say about the original mass function from this?